10 grams of AlCl3 contains a certain number of moles in particles.
1 mole of AlCl3 weighs (26.9815386+(3*35.453)=133.3405386), let's round that to 133.3405 grams.
So 10 grams = 10/133.3405 = 0.075 mole of AlCl3 (rounded), but for each AlCl3 we have, you get 4 ions (1 Al3+, 3Cl-). So we have 0.075 moles, but we get 4 times as many ions. 0.075*4=0.3 mole
1 mole = 6.022141x1023 (Avogadro's Number)
0.3 mole = 1.81x1023 ions
1 mole of AlCl3 will dissociate into 4 moles of ions in aqueous solution: 1 mole of Al+3 ions and 3 moles of Cl- ions.
The van't Hoff factor for AlCl3 is 4. AlCl3 dissociates into Al3+ and 3Cl- ions when it dissolves in water, resulting in a total of 4 ions in solution per formula unit of AlCl3.
0.355 M AlCl3 (3 moles Cl/1 mole AlCl3)= 1.07 M Cl================Naturally.0.355 M Al============add1.43 M total=========
AlCl3 can precipitate with compounds that contain chloride ions, such as NaCl, KCl, or HCl, to form insoluble AlCl3 complexes. This reaction can be used for the precipitation of aluminum ions from a solution.
The pH of a 0.1M solution of AlCl3 would be quite low due to the hydrolysis of aluminum ions in water, leading to the formation of H+ ions and acidic conditions. The exact pH value would depend on the equilibrium constants for the hydrolysis reactions involving aluminum ions.
In 1 mol of AlCl3, there are 3 chloride ions. First calculate the moles of AlCl3 in the solution: 65.5 mL is 0.0655 L. Multiply 0.0655 L by 0.210 mol/L to get the moles of AlCl3. Finally, multiply this by 3 to find the number of chloride ions in the solution.
In one mole of AlCl3, there are 3 moles of chloride ions. The molar mass of AlCl3 is 133.34 g/mol. Therefore, there are 6.022 x 10^23 chloride ions in 133.34 g of AlCl3. To find the number of chloride ions in 400 g of AlCl3, you would calculate (6.022 x 10^23) x (400 g / 133.34 g/mol) = approximately 1.8 x 10^24 chloride ions.
In 6.75 moles of AlCl3, there are 6.75 times Avogadro's number of formula units, which is approximately 4.07 x 10^24 formula units. Each formula unit of AlCl3 contains one Al^3+ ion and three Cl^- ions. Therefore, in 6.75 moles of AlCl3, there are also 4.07 x 10^24 Al^3+ ions and 1.22 x 10^25 Cl^- ions.
1 mole of AlCl3 will dissociate into 4 moles of ions in aqueous solution: 1 mole of Al+3 ions and 3 moles of Cl- ions.
The van't Hoff factor for AlCl3 is 4. AlCl3 dissociates into Al3+ and 3Cl- ions when it dissolves in water, resulting in a total of 4 ions in solution per formula unit of AlCl3.
0.355 M AlCl3 (3 moles Cl/1 mole AlCl3)= 1.07 M Cl================Naturally.0.355 M Al============add1.43 M total=========
AlCl3 can precipitate with compounds that contain chloride ions, such as NaCl, KCl, or HCl, to form insoluble AlCl3 complexes. This reaction can be used for the precipitation of aluminum ions from a solution.
3 mol Cl- ions per mol AlCl3 , so 3 * 0.5 = 1.5 mol/L Cl-
It depends on what it is reacting with.
The pH of a 0.1M solution of AlCl3 would be quite low due to the hydrolysis of aluminum ions in water, leading to the formation of H+ ions and acidic conditions. The exact pH value would depend on the equilibrium constants for the hydrolysis reactions involving aluminum ions.
The formula for aluminum chloride is AlCl3, which consists of aluminum ions (Al3+) and chloride ions (Cl-). In the compound, aluminum donates three electrons to chloride, resulting in the formation of three chloride ions for every aluminum ion.
1 mol of AlCl3 has mass 133.5 g and has 3 mol Cl- atoms so 400.5 g AlCl3 means 3 mol AlCl3 which means 9 mol Cl- atoms. So Ans = 9 X 6.022 X 1023 = 5.4198 X 1024 Cl- atoms